2006 AMC 12A Problems/Problem 23
Contents
Problem
Given a finite sequence of real numbers, let be the sequence of real numbers. Define and, for each integer , , define . Suppose , and let . If , then what is ?
Solution 1
In general, such that has terms. Specifically, To find x, we need only solve the equation . Algebra yields .
Solution 2
For every sequence of at least three terms,
Thus for , the coefficients of the terms in the numerator of are the binomial coefficients , and the denominator is . Because for all integers , the coefficients of the terms in the numerators of are for . The definition implies that the denominator of each term in is . For the given sequence, the sole term in is Therefore, so , and because , we have .
See also
2006 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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